The h Bar - 2018-01-16 (page 1 of 4)

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Anonymous

1:32 AM

@Semiclassical If you happen to come around, please have a look at this question: physics.stackexchange.com/questions/380214/…

JAVA

coding . . . . . . JAVA. . . . . .

Listening to pandora.com/artist/ingrid-michaelson/girls-chase-boys-single/…

1 hour later…

2:49 AM

@Slereah I think, in eq. I.75, Scherk takes the $x^{\mu}$ from I.55 $x^{\mu}(\tau,\sigma) = q_0^{\mu} + 2 \alpha' p_0^{\mu} \tau + i \sqrt{2 \alpha'} \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}[a_n^{\mu}(0)e^{-int} - a_n^{\mu *}(0)e^{int}] \cos(n \sigma) $ and then wants to factor out $2 \alpha' p_0^{\mu}$ so that $x^{\mu}$ becomes

$x^{\mu}(\tau,\sigma) = 2 \alpha' p_0^{\mu} [\frac{q_0^{\mu}}{2 \alpha' p_0^{\mu}} + \tau + i \sqrt{2 \alpha'} \sum_{n=1}^{\infty} \frac{1}{2 \alpha' p_0^{\mu}\sqrt{n}}[a_n^{\mu}(0)e^{-int} - a_n^{\mu *}(0)e^{int}] \cos(n \sigma) ] = 2 \alpha' p_0^{\mu} \tau'$ (which amounts to a change of variables possible by conformal invariance), but because this makes no sense (divided by a vector $p^{\mu}$) he forms a dot product with some vector $n^{\mu}$ to do this

Bernardo Meurer

@dmckee @JohnRennie Another one :)

Ahh, that's a vector, that explains why lightcone coordinates actually pop up, I.88

No wait, not dividing by a vector...

Seems like Scherk shows the fields $x^{\mu}$ in the NG action satisfy the wave equation, and then in section 6 (eq. I.66 on) shows the coordinates $\tau,\sigma$ also satisfy the same wave equation, so that we can set one of our $x^{\mu}$ fields equal to one of the $\tau,\sigma$ coordinates, say $\tau$, and so for some reason, maybe to guarantee we have a time-like coordinate, dot-products $x^{\mu}$ with a time-like vector so he can do the stuff I posted above, hmm

Something about the $n$ vector fixing the parametrization...

3 hours later…

5:49 AM

@JohnRennie isn’t here

I was going to tell him something

Oh well

1 hour later…

Bernardo Meurer

6:55 AM

I can't stop watching John Duffield being teleported into a CGI spaceship

Captain Bohemian

It's warm outside and it still feels so cold indoors. I was forced to stop cold shower due to feeling over cold.

@BernardoMeurer are you sure it's CGI

He might have an actual spaceship

Bernardo Meurer

@Slereah Good question, now I don't know

Maybe that's why @0celo7 has been seeing manifolds in the sky

7:10 AM

I wonder if he'll come back

It's been a while since he dropped by

Bernardo Meurer

That is true

I miss him to be honest

Hmm, it turns out MSVC doesn't implement the getline function ...

Bernardo Meurer

@JohnRennie But that was standardized in POSIX-2008

*POSIX.1-2008

It implements std::getline

Bernardo Meurer

That's c++ mumbo-jumbo

get it away from me

7:21 AM

But not the straightforward C getline function

I must admit that surprises me

C++ doesn't implement all of the C standard

just most of it

Bernardo Meurer

@Slereah Well, to be fair, getline isn't in the C standard IMHO

It's not C99

And no one cares for C11

Also, it was a GNU extension before POSIX adopted

8:07 AM

Hello my dudes

where are the diff geo people

I am in need of the diff geo

Captain Bohemian

go to math chat

really? I found it's like there are many math conversnt people in math chatroom. Or maybe compared to you, they are deficient.

Nah, I am just kidding

8:29 AM

→ 1 message moved to trash

Silently the ninja room owner stalks his prey

boo

8:46 AM

"The background philosophy behind this paper is very simple: Finsler metrics are special pseudo-Riemannian metrics in a special vector bundle."

very simple indeed

Captain Bohemian

9:18 AM

do spin networks represent real physical interactions, like the case of Feymann diagrams? Or are they just imagined graphs covering the spacetime manifold in order to describe the spacetime manifold, like coordinate patches?

From what I can remember, spin networks represent space at an instant

They evolve according to some dynamics

Captain Bohemian

so are they real physical interactions or imagined graphs on that space?

what does "real" mean

It's the quantum state of the $3$-metric

you may decide how real it is

Captain Bohemian

doesn't each Feymann diagram reprsent a real physical interaction, like an electron and a positron merge into two photons?

But if spin networks are just like coordinates covering a manifold, they are not real; they are the grids people carve on the manifold to describe the manifold.

hey physicists

why is symplectic geometry important to you? tell me

tell me all

Captain Bohemian

9:29 AM

it can describe the dynamics of geometry

that sounds nonspecific. please elaborate?

@BalarkaSen The Hamiltonian

tell me

tell me more

Did she put up a fight

Captain Bohemian

@Slereah quantum state of the $3$-metric? So they are the real interactions? These edges in spin networks represent real particles with spins labeled by those edges.

9:35 AM

it's gives you the fancy version of the Legendre transform basically?

Well no, quantum states aren't interactions

You have to evolve spin networks to get the dynamic

Spin networks are eigenstates of an operator associated to the area of a region of space

Captain Bohemian

really? it turns out spin networks really have an operator to give rise to. Last time I asked you this question, but you didn't reply me; probably you were sleeping.

yes, sometimes I do that

the human sleep

Captain Bohemian

symplectic geometry can be performed canonical transformations.

yes

that way you can do that trick of turning Lagrange equations into first order equations

and do a nice split of time and space, if in spacetime

although of course you need the horrible polysymplectic geometry if you're using fields

user image

Well summarized

9:51 AM

rip

rip in diagrams

also why is that diagram lopsided

the bundle is lopsided

Captain Bohemian

so those edges in a spin network represent quantum states of 3-metric? then why does a vertice in a spin network represent?

No, the whole diagram is the state

Captain Bohemian

a spin network represents a quantum state?

yes

you sum over spin networks to get the propagator

Captain Bohemian

10:01 AM

so a vertice in a spin network represents tensor product (entanglement) between particles?

10:14 AM

the loops are apparently holonomies of the $3$-connection

Captain Bohemian

how are loops related to spin networks?

the spin network is made of loops

[Random]

New thoughts on changing history:

(model independent(?))

Start at the present day, travel x hours back to the past

When the time traveler depart from the present as seen in the frame of the laboratory in the present day, and go back to the past, the worldline between x hours ago and the present as seen in this frame (and relativistically in other moving frames) will be updated instantly at all events along the worldline.

It's as if the hamiltonian that gives the evolution of all the events along the worldline changes into a new one at all events in this worldline the moment the time traveler arrived back in its past as seen from the laboratory frame.

Need to check later whether all of this makes sense...

(NB, this is not GR, as one cannot change spacetime topologies on the fly)

10:30 AM

@JohnRennie Last question? I am unable to figure out how these two molecules are mirror images...

@Abcd Here or in problem solving?

problem solving.

11:27 AM

Maybe I should read Cartan to get a good grasp on connections

He is the French

11:52 AM

"Let $\gamma : I \to M$ be a path with $\gamma(0) = p$ and $\gamma(1) = q$. For every $v \in T_pM$ there exists a unique horizontal lift $\tilde \gamma$ such that $\tilde \gamma(0) = v$ and $\tilde \gamma(1) \in T_q$"

What

I thought the geodesics weren't unique between two points for general manifolds

Or is it true

I'm not sure

I guess it might be with a specified tangent vector $v$?

12:17 PM

Oh wait

It was a RUSE THEOREM

The real theorem is the same but also requires the connection to be uniformly vertically bounded

12:48 PM

@ACuriousMind How does the connection work, like, formally speaking

I'm guessing you need to define a basis of the horizontal subspace of $TE$

Is this related to the connection form

Also since this is just $\Bbb R^n$ how does rotation of the basis work for a general connection

does it correspond to anything

Future Historian

So, about that nuke explosion question, I wonder if a 10 km long, 6 km spacecraft could realistically survive that.

1:03 PM

@Slereah the horizontal distribution is the kernel of the form

So the connection form is just the normal form to the connection?

Wot

Future Historian

Gentlemen?

In theory, could a carbon nanotube-tungsten alloy be possible?

Or is that not really doable?

@0celo7 If I have some vector $h \in H_e E$, and a connection form $\omega$, is $\omega_{abc...} h^a = 0$

Oh wait, I guess the connection form isn't rly a form

Future Historian

@Slereah? Question: is it possible that a carbon nanotube-tungsten alloy could work in real life?

1:17 PM

I dunno man

Future Historian

Mainly for vehicle and spacecraft armour?

@0celo7? Let me guess: carbon nanotube-tungsten alloy = unrealistic?

Or.....is current science able to allow that (in theory)?

Wot

He wants to know if he can make a Very Strong Tank

Future Historian

Or in this case, a very strong military spacecraft.

:P

I don't think any of us are really big in material science

1:25 PM

@0celo7 what's a simple definition of the connection form for the Ehresmann connection

That isn't "an endomorphism of the bundle"

Oh wait

I think ironically nlab's definition is more palatable

"A Cartan-Ehresmann connection on P P is a Lie algebra-valued 1-form"

But that's only for principal bundles

If you want a really general theory, check Kolar, Michor, and Slovak, Natural Operations in Differential Geometry.

Also I'm sure the new book by Loring Tu is good.

I'll probably buy it one of these days.

amazon.com/…

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thx

I'll give it the old lookaroo

I'm pretty glad to live in the internet age, really

Can you imagine having to go to the library every time you have to look up a thing

And if the book isn't available, wait a few weeks for the library to order it

@Slereah If you want something just for vector bundles, I can recommend Metric Structures in Differential Geometry by G. Walschap.

1:35 PM

Thanks

@Slereah The best explanation of associated affine connections seems to be in Bishop and Crittenden, Geometry of Manifolds.

Is the connection $1$-form an actual one form of $\Lambda^1 E$ btw

That's kind of a historical book but is very important for Riemannian geometers.

You know I tried reading that quantum logic thing

and I still don't know what it's for

The basic idea seems to be that you have structures about Hilbert subspaces which are similar to boolean algebras

Except it doesn't commute for everything

I always think about the connection as the exterior covariant derivative

1:38 PM

but I'm not sure what it's for

go away

@Slereah The connection form is a $\mathfrak{gl}(n)$-valued 1-form.

No you

@Slereah If you have a vector bundle you can take it's frame bundle, which is a principle GL(n)-bundle, isn't it?

Then your notion of a connection 1-form is well defined

It's a gl(n)-valued 1-form like 0celo7 said

@0celo7 What if it's not on $TM$

Any vector bundle has it's structure group in GL(n)

Is it generally valued as the structure group of the bundle under consideration?

1:42 PM

Well, GL(r) depending on the rank of the bundle

Is what I'm asking

(In case it's some gauge business)

@Slereah The way you should think about it is: vector bundles: distribution <-> covariant derivative, principal bundles: distribution <-> connection form

that is to say, the connection form is less important for vector bundles

but for a vector bundle, the connection is a section of $\mathfrak{gl}(r)\otimes T^*M$, restricted to a coordinate domain

Alright

never really liked the theory of moving frames myself

Ted is judging you

1:49 PM

I guess it's like $\omega^\mu_a \partial_\mu de^a$, for some basis $de$ of $gl$ or something?

wot

@0celo7 differential forms are too hard

Well similarly to gauge fields

$A^\mu_a \partial_\mu \tau^a$

@BalarkaSen it gets hard when there's submanifolds involved

i like pictures

1:50 PM

With $\tau^a$ the basis of the Lie group

@Slereah that looks like a vector, not a 1-form

Well $A_\mu^a$

If you prefer

ok

gotta make food

running late af

l8er

Baby Cakes - Diary #3

2

this is poetry of the modern age

1:55 PM

I think you mean the age of 10 years ago

@Slereah Check out Bishop and Crittenden

Also check out Michor's book on diff geo, the draft is free on his website

Brad Neely became a huge meme, then fell back into obscurity, then got his own TV show and then fell back into obscurity again

I always think I would be a better person if I learned diff geo from him

Oh man

Why are there so many math books

I just want to go back to Ancient Greece

to confuse readers

1:56 PM

where you just had to go to see other people for math

And that was all there was

@Slereah damn i don't know about this

I don't even want the library of Alexandria to exist

also Spivak of course

China, IL (meaning China, Illinois) is an American animated television series created by Brad Neely for the Adult Swim programming block on Cartoon Network. The series takes place at The University of China, Illinois, dubbed the "Worst College in America" and located at the edge of town. The school's poor reputation is celebrated by the school's uncaring faculty and staff, constantly shown drinking while teaching and/or trying to avoid teaching altogether. The series was originally conceived as a web series on Adult Swim's defunct comedy website, Super Deluxe, in 2008. Neely, who had done Baby...

The show he made

It's basically a continuation of his old videos

2

Q: How do connection 1-form and Ehresmann version of connections relate to each other?

I am studying connections on abstract manifolds. So far, I have read several equivalent definitions but I can't establish the equivalence between them on my own. The first definition is the Ehresmann connection that defines a connection on a manifold as a distribution of vector spaces completing...

differential-geometry

Might be relevant to what I want

The pullback bundle $\gamma^* E$ is just the bundle over the curve, right?

yep

2:10 PM

Aight

Or at least the projection $g : \gamma^* E \to E$ has $\operatorname{Im}(g)$ being that, I suppose

2:50 PM

@JohnRennie Did you once tell me that a collapsing universe need not be finite?

Or am I losing it?

Sure

You can have collapsing $\Bbb R^n$

The topology of the singularity is the same topology as the spacelike hypersurface itself

but geometrically it's just a point because the metric is just $g = 0$

Well what's this about quintessence?

Could anyone tell me how I could quickly prove that the Weyl tensor in d=3 dimensions vanishes?

Should I be using some sort of symmetry properties for one o fthe indices in$C_{\alpha\beta\mu\nu}$?

Since it has the same symmetries as Riemann, that would mean R also vanishes but it doesn't... leading me to believe it's something else I'm overlooking.

@SirCumference what do you mean

in 3 dimensions, the Riemann tensor only depends on the Ricci tensor

Hence the contribution of the Weyl tensor vanishes

Which you can prove indeed with the properties of the Riemann tensor

I...don't know the differential geometry for this

2:58 PM

By checking that with just the indices $1,2,3$, the components of the Weyl tensor vanish by exchanging them

Welp, can our universe, given what we know, be infinite yet collapse?

I think it's basically showing that any element is gonna be of the form 1231 or worse

And that will mean that the components are $C = -C$ sort of thing

$$ C_{abcd}^{}=-C_{bacd}=-C_{abdc}$$

Bro...I just want to know the answer to that one question

If an element is of the form $C_{aacd}$ then it is equal to $-C_{aacd}$

That kind of stuff

@Slereah Sure, but wouldn't the same hold for $R_{\alpha\beta\mu\nu}$?

3:02 PM

I dont...

Ok

Which doesn't vanish in d=3?

@user55789 Not quite

Or does it and am I saying something stupid..

I don't think the Riemann tensor has exactly the same symmetries?

hm, it does, at least for the interchange

My textbook says

3:04 PM

Isn't it the Binachi identity

"$C_{\rho\sigma\mu\nu}$ is designed so that all possible contractions vanish, while it retains the symmetries of the Riemann tensor"

Specifically they note

The Weyl tensor also obeys the Bianchi identity

Oh, you're talking about skew symmetry

\begin{equation}
C_{\rho\sigma\mu\nu} = C_{[\rho\sigma][\mu\nu]}
\end{equation}

$R_{abcd} = -R_{bacd} = -R_{abdc}$

3:05 PM

\begin{equation}
C_{\rho\sigma\mu\nu} = C_{\mu\nu\rho\sigma}
\end{equation}

yeah might be this

\begin{equation}
C_{\rho[\sigma\mu\nu]} =0
\end{equation}

Just check what you obtain for $C_{aabc}$, $C_{abbc}$ and $C_{abcc}$

Those are the only possible combinations

@user55789 You can just use $ instead of \begin...

Or something

Or the double dollar sign

3:06 PM

Well the problem is that this line of reasoning leads to $R$ also being zero, no?

for extra fancy equations

Double dollar is bad practice if you're a $\LaTeX$ whore like me

it's great for chat though

^

I also use \Bbb on chat

even though it's deprecated

$\Bbb{VERY BAD}$

3:08 PM

Disgusting.

:)

$\mathfrak{VERBOTTEN}$

You are smoothly deforming the conversation of the subject of this conversation from the task at hand...

Namely, identifying why the symmetry argument does not hold for Riemann while it does for Weyl

I'm guessing the proof involves one of the symmetry that doesn't hold for the Riemann tensor

physicsforums.com/threads/…

Carroll argues that all symmetries are inherited...

Confuzzling.

The extra property of Weyl is $C^{i}{}_{jki}=0$

Not sure if it helps

Oh wait

If $C^{i}{}_{jki}=0$

Maybe what you need to do is recast every other component as $C^{i}{}_{jki}$

And since two components will always be the same

that is plausible

Like $C^{i}{}_{ijk}= - C^{i}{}_{jki} = 0$ or something

the hard part is having a general argument for every possible combination

Although I guess that's not quite the same thing

3:23 PM

@Slereah Wait is that actually him

since $C^{i}{}_{ijk} = C^{1}{}_{1jk} + C^{2}{}_{2jk} + C^{3}{}_{3jk}$

@SirCumference yes

are you messing with me

How many John Duffields do you know that ramble on about relativity

Maybe try to first check out how many components of the tensor are actually independant

IIRC it's not that many

Then show that these components are 0

Actually the number of independant components has a factor of $N - 3$

ion.uwinnipeg.ca/~vincent/4500.6-001/Cosmology/WeylTensor.htm

^if you can find the formula used there then you can just do the definition of the Weyl tensor to find it's 0

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